Problem 15 - Applied Calculus

Question

Use integration by parts to find each integral. \(\int \sqrt{x} \ln x d x\)

Step-by-Step Solution

Answer

The integral is \( \int \sqrt{x} \ln x \, dx = \frac{2}{3} x^{3/2} \ln x - \frac{4}{9} x^{3/2} + C \).

Step 1: Identify Parts for Integration by Parts

We know the formula for integration by parts is \( \int u \, dv = uv - \int v \, du \). First, we need to decide which parts of the integral will be \( u \) and \( dv \). Let's choose \( u = \ln x \) because its derivative simplifies the expression, and \( dv = \sqrt{x} \, dx \) so that we can integrate it easily.

Step 2: Find \( du \) and \( v \)

Differentiate \( u \) to find \( du \): \( du = \frac{1}{x} \, dx \). Integrate \( dv \) to find \( v \):\[ v = \int \sqrt{x} \, dx = \frac{2}{3} x^{3/2} \].

Step 3: Apply the Integration by Parts Formula

Substitute \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula:\[ \int \sqrt{x} \ln x \, dx = \ln x \cdot \frac{2}{3} x^{3/2} - \int \left( \frac{2}{3} x^{3/2} \right) \left( \frac{1}{x} \right) dx \].

Step 4: Simplify and Integrate the Remaining Integral

Simplify the remaining integral:\[ \int \frac{2}{3} x^{3/2-1} \, dx = \frac{2}{3} \int x^{1/2} \, dx \]. Now integrate this:\[ \frac{2}{3} \times \frac{2}{3} x^{3/2} = \frac{4}{9} x^{3/2} \].

Step 5: Combine the Results

Combine the results from Step 3 and Step 4:\[ \int \sqrt{x} \ln x \, dx = \frac{2}{3} x^{3/2} \ln x - \frac{4}{9} x^{3/2} + C \], where \( C \) is the constant of integration.

Key Concepts

definite integral

A definite integral is a fundamental concept in calculus that helps determine the exact area under a curve between two specified points, typically along the x-axis. Unlike indefinite integrals, which include a constant of integration, definite integrals have fixed upper and lower limits. This gives them a specific numerical value. To compute a definite integral, you first find the antiderivative of the function and then evaluate it at the upper and lower limits. The result is the difference between the values at these two points.
Here's a short list of steps involved:

• Identify the function to integrate from the given interval.
• Compute the antiderivative without integrating the limits.
• Substitute the upper limit into the antiderivative and calculate the value.
• Subtract the value obtained when substituting the lower limit from the upper limit's value.

Using the definite integral, you can solve real-world problems involving areas, volumes, and total accumulations, providing a powerful tool for analysis.

calculus problem solving

Calculus problem solving involves a set of techniques and methods to tackle different types of calculus problems systematically. This can range from simple derivatives to more complex integrals and differential equations. A solid understanding of calculus concepts and formulas is crucial for solving problems effectively. The integration by parts technique, demonstrated in the original solution, is a classic problem-solving approach in calculus, especially when dealing with products of functions. In integration by parts, the choice of functions to differentiate and integrate is essential. Typically, logarithmic and algebraic functions are chosen for differentiation, while polynomials or exponential functions are better candidates for integration.
Here are some general steps for effective calculus problem solving:

• Understand the problem and identify the type of calculus operation needed.
• Simplify the problem if possible, such as breaking down the equation or expression.
• Use calculus techniques like substitution, integration by parts, or partial fractions as needed.
• Verify your solution by differentiating or using graphs to check its correctness.

This structured approach not only aids in finding a solution but also hones critical thinking skills in mathematical problem-solving.

integral calculus

Integral calculus is a branch of calculus focused on finding the total or accumulated quantity, such as the area under a curve, the length of a curve, or the accumulation of change. It deals with integrals and the properties of integral functions. Integrals are classified into two major types: definite integrals, which compute the value between limits, and indefinite integrals, which denote a set of functions differing by a constant. The fundamental theorem of calculus establishes the connection between differentiation and integration, ensuring that derivatives and integrals are inverse operations. This theorem has two main components:

• The first part states that the integral of a function can be reversed by differentiating.
• The second part provides a practical way to compute a definite integral using its antiderivative.

Integral calculus is crucial in fields such as physics, engineering, and economics, where calculating things like velocity, area, and accumulated cost or interest is essential. Understanding integral calculus concepts aids in solving real-world problems, involving continuous change or accumulation.